9.2 Determinants, Matrix Inverses, and Cramer's Rule

9.2 Determinants, Matrix Inverses, and Cramer's Rule

The determinant is an indispensable scalar invariant associated with square matrices. On the GRE Mathematics Subject Test, questions frequently test axiomatic determinant properties, efficient computation via row operations or block decompositions, the classical adjugate inverse formula, and Cramer's Rule for isolated variables.

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Axiomatic Characterization and the Leibniz Formula

For a matrix $A \in M_{n \times n}(\mathbb{R})$, the determinant is the unique function $\det: M_{n \times n}(\mathbb{R}) \to \mathbb{R}$ satisfying three foundational axioms on its rows $\mathbf{r}_1, \dots, \mathbf{r}_n$:

1. Multilinearity: $\det$ is linear in each row vector when all other rows remain fixed: $\det \begin{pmatrix} \mathbf{r}_1 \\ \vdots \\ c\mathbf{u} + d\mathbf{v} \\ \vdots \\ \mathbf{r}_n \end{pmatrix} = c \det \begin{pmatrix} \mathbf{r}_1 \\ \vdots \\ \mathbf{u} \\ \vdots \\ \mathbf{r}_n \end{pmatrix} + d \det \begin{pmatrix} \mathbf{r}_1 \\ \vdots \\ \mathbf{v} \\ \vdots \\ \mathbf{r}_n \end{pmatrix}.$
2. Alternating Property: If two rows of $A$ are interchanged, the determinant reverses sign. Consequently, if $A$ contains two identical rows, $\det(A) = 0$.
3. Normalization: $\det(I_n) = 1$.

These axioms uniquely imply the Leibniz expansion formula over the symmetric group $S_n$: $\det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i, \sigma(i)},$ where $\operatorname{sgn}(\sigma) \in {+1, -1}$ denotes the parity (signature) of the permutation $\sigma$.

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Essential Operational and Multiplicative Properties

Rigorous application of the following multiplicative identities resolves many theoretical GRE problems instantaneously:

• Multiplicativity: $\det(AB) = \det(A)\det(B)$. By induction, $\det(A^k) = (\det A)^k$.
• Transpose Invariance: $\det(A^T) = \det(A)$. All properties holding for rows apply identically to columns.
• Homogeneity of Degree $n$: For any scalar $c \in \mathbb{R}$ and $n \times n$ matrix $A$: $\det(cA) = c^n \det(A).$
• Invertibility Criterion: A matrix $A$ is invertible if and only if $\det(A) \neq 0$. When invertible: $\det(A^{-1}) = \frac{1}{\det(A)}.$
• Block Triangular Matrices: If $A$ is $k \times k$ and $C$ is $(n-k) \times (n-k)$: $\det \begin{pmatrix} A & B \\ 0 & C \end{pmatrix} = \det(A) \det(C).$
• Schur Complement: If $A$ is invertible, the block matrix determinant expands as: $\det \begin{pmatrix} A & B \\ C & D \end{pmatrix} = \det(A) \det(D - C A^{-1} B).$

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Determinants via Elementary Row Operations

Computing determinants by definition is computationally prohibitive for $n \ge 4$. Gaussian row reduction simplifies $A$ to an upper triangular matrix $U$:

• Row Swap ($R_i \leftrightarrow R_j$): Multiplies the determinant by $-1$.
• Row Scaling ($c R_i \to R_i$): Multiplies the determinant by $c$.
• Row Addition ($R_i + c R_j \to R_i$): Leaves the determinant invariant.

Once in triangular form, the determinant is the product of diagonal entries: $\det(U) = \prod_{i=1}^n u_{ii}$.

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Cofactor Expansion and the Classical Adjugate

Let $M_{ij}$ denote the $(n-1) \times (n-1)$ submatrix obtained by deleting row $i$ and column $j$ of $A$. The $(i, j)$-cofactor is: $C_{ij} = (-1)^{i+j} \det(M_{ij}).$ Laplace expansion evaluates $\det(A)$ along any arbitrary row $i$ or column $j$: $\det(A) = \sum_{k=1}^n a_{ik} C_{ik} = \sum_{k=1}^n a_{kj} C_{kj}.$

The Adjugate Matrix Formula

The adjugate (or classical adjoint), denoted $\operatorname{adj}(A)$, is the transpose of the cofactor matrix: $(\operatorname{adj}(A))_{ij} = C_{ji}.$ The fundamental adjugate identity states: $A \operatorname{adj}(A) = \operatorname{adj}(A) A = (\det A) I_n.$ When $\det(A) \neq 0$, this yields the explicit matrix inverse formula: $A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A).$

Properties of the Adjugate

• $\det(\operatorname{adj}(A)) = (\det A)^{n-1}$.
• For $n \ge 3$, $\operatorname{adj}(\operatorname{adj}(A)) = (\det A)^{n-2} A$.

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Cramer's Rule and Geometric Interpretation

Cramer's Rule

For a system $A\mathbf{x} = \mathbf{b}$ with $A \in M_{n \times n}(\mathbb{R})$ and $\det(A) \neq 0$, the unique solution components are: $x_i = \frac{\det(A_i)}{\det(A)},$ where $A_i$ is the matrix formed by replacing the $i$-th column of $A$ with $\mathbf{b}$. Cramer's Rule is especially powerful on the GRE when asked for only a single unknown, avoiding a full matrix inversion.

Geometric Interpretation: Hyper-Volume Scaling

The absolute value $|\det(A)|$ represents the $n$-dimensional volume of the parallelotope formed by the columns of $A$. Under the linear transformation $T(\mathbf{x}) = A\mathbf{x}$, the volume of any measurable set $S \subset \mathbb{R}^n$ scales as $\operatorname{Vol}(T(S)) = |\det(A)| \operatorname{Vol}(S)$. The sign of $\det(A)$ specifies whether the transformation preserves ($+1$) or reverses ($-1$) spatial orientation.

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Step-by-Step Worked Problems

Problem 1: Determinant with Adjugate and Scaled Inverses

Let $A$ be a $3 \times 3$ real matrix with $\det(A) = 4$. Compute $\det(2 A^{-1} \operatorname{adj}(A^T))$.

Solution:

1. Apply homogeneity of degree $n = 3$ to pull the scalar $2$ outside: $\det(2 M) = 2^3 \det(M) = 8 \det(M).$
2. Use the multiplicative property of the determinant: $\det(A^{-1} \operatorname{adj}(A^T)) = \det(A^{-1}) \det(\operatorname{adj}(A^T)).$
3. Evaluate individual factors: $\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{4}.$ $\det(\operatorname{adj}(A^T)) = (\det(A^T))^{n-1} = (\det A)^{3-1} = 4^2 = 16.$
4. Combine all factors: $\det(2 A^{-1} \operatorname{adj}(A^T)) = 8 \times \left( \frac{1}{4} \times 16 \right) = 8 \times 4 = 32.$

Problem 2: Cramer's Rule for a Single Variable

Solve for $y$ in the system: $\begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 7 \end{pmatrix}$

Solution:

1. Compute $\det(A)$: $\det(A) = 2(4 - 1) - 1(2 - 1) + 1(1 - 2) = 2(3) - 1(1) + 1(-1) = 6 - 1 - 1 = 4.$
2. Form $A_2$ by replacing the second column with $\mathbf{b} = (4, 1, 7)^T$: $A_2 = \begin{pmatrix} 2 & 4 & 1 \\ 1 & 1 & 1 \\ 1 & 7 & 2 \end{pmatrix}$
3. Expand $\det(A_2)$ along the first row: $\det(A_2) = 2(2 - 7) - 4(2 - 1) + 1(7 - 1) = 2(-5) - 4(1) + 1(6) = -10 - 4 + 6 = -8.$
4. Apply Cramer's Rule: $y = \frac{\det(A_2)}{\det(A)} = \frac{-8}{4} = -2.$

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GRE Exam Traps & Pitfalls

Trap 1: Dropping the Exponent on Scalar Scaling Writing $\det(cA) = c \det(A)$ is the single most common determinant blunder. For an $n \times n$ matrix, every row is multiplied by $c$, yielding $\det(cA) = c^n \det(A)$.

Trap 2: Assuming Linearity Across Addition In general, $\det(A + B) \neq \det(A) + \det(B)$. For instance, $\det(I_2 + (-I_2)) = \det(0) = 0$, but $\det(I_2) + \det(-I_2) = 1 + (-1)^2 = 2$.

Trap 3: Forgetting the Transpose in the Adjugate Definition The adjugate is the transpose of the cofactor matrix: $(\operatorname{adj}(A)){ij} = C{ji}$. If you calculate cofactors and fail to transpose before multiplying, the result will not equal $(\det A) I_n$ unless the matrix is symmetric.